Transformers Questions and Answers

Answer: B

Laminating the core minimizes Eddy Current loss.

The alternating flux induces circulating (eddy) currents in the core. Building the core from thin laminations insulated from each other breaks up the current paths, and since eddy loss is proportional to the square of the lamination thickness, thinner sheets cut it sharply.

Hysteresis loss depends on the magnetic material and the flux density, and is reduced by using silicon steel, not by laminating. 'Heat loss' is not a separate kind of loss; every loss ends up as heat. So 'All of the above' is wrong too.

Answer: A

Iron loss occurs in the Core.

Iron (core) loss is the sum of hysteresis loss and eddy current loss, both caused by the alternating flux in the magnetic core. It depends on the voltage and frequency, and is nearly constant from no load to full load.

Loss in the windings is copper loss, \(I^{2}R\). The insulating oil carries heat away but is not where iron loss arises, and the tank (main body) carries only small stray losses.

Answer: C

On no load, the negligible loss is Copper loss.

With the secondary open, the primary draws only the small no-load current, a few per cent of the rated current. Copper loss is proportional to the square of the current, so it is tiny.

Hysteresis and eddy current losses depend on the flux, which is set by the applied voltage and is the same on no load as on full load. They are therefore present in full, which is why the open circuit test measures iron loss. The losses are clearly not all equal.

Answer: B

Copper loss occurs in the Winding.

Copper loss is the \(I^{2}R\) loss in the resistance of the primary and secondary windings: \(P_{cu} = I_{1}^{2}R_{1} + I_{2}^{2}R_{2}\). It varies with the square of the load current.

The core is where iron loss (hysteresis and eddy current) occurs. The main body and the bushings carry only small stray losses and are not where copper loss arises.

Answer: C

Efficiency is maximum when Copper loss = Iron loss.

Efficiency is \(\eta = \frac{xS\cos\phi}{xS\cos\phi + P_{i} + x^{2}P_{cu}}\), where \(x\) is the fraction of full load. Iron loss \(P_{i}\) is constant, and copper loss grows as \(x^{2}\). Setting \(d\eta/dx = 0\) gives \(x^{2}P_{cu} = P_{i}\): the variable loss equals the constant loss.

Hysteresis and eddy losses are both parts of the constant iron loss, and core loss is the same thing as iron loss, so neither of those equalities is a condition. At maximum efficiency total loss is twice the copper loss, not two-thirds of it.

Answer: A

Eddy current loss is reduced by Reducing the thickness of laminations.

Eddy current loss per unit volume varies as \(B_{max}^{2}f^{2}t^{2}\), where \(t\) is the lamination thickness. Thinner laminations confine each eddy current path to a narrower strip with higher resistance, and the loss falls with the square of the thickness.

Thicker laminations increase the loss. The air gap changes the reluctance and therefore the magnetizing current, not the eddy current loss in the laminations.

Answer: C

Coupling is increased by Using the magnetic core of low reluctance.

Coupling is the fraction of the primary's flux that also links the secondary. A core of low reluctance, a high-permeability closed iron path, carries most of the flux through both windings and so reduces leakage flux.

More turns raise the flux linkage of each winding but do not make the flux that links one winding link the other. The material of the windings has nothing to do with magnetic coupling, and the quality of the oil affects insulation and cooling only.

Answer: C

The loss that varies with load is Copper loss.

Copper loss is \(I^{2}R\) in the windings, so it changes with the square of the load current: at half load it is a quarter of its full-load value.

Hysteresis loss and eddy current loss together make up the core (iron) loss, which depends on the voltage and frequency, not on the load. With the supply voltage constant, all three of those options stay essentially constant from no load to full load.

Answer: D

The power factor in a transformer Depends on the power factor of the load.

A transformer passes power from one circuit to another and draws only a small magnetizing current of its own. On load, the primary current is set mainly by the secondary current, so the primary power factor follows the power factor of whatever load is connected.

It is not always unity, leading or lagging: a resistive load gives nearly unity, an inductive load lagging and a capacitive load leading. Only on no load is it fixed, and then it is low and lagging.

Answer: A

At no load the current Lags behind the applied voltage by 80°.

The no-load current has two components: a large magnetizing component that lags the voltage by 90° and a small active component in phase with it, which supplies the core loss. Because the magnetizing part dominates, the no-load power factor is low, about 0.1 to 0.2, so the current lags by roughly 80°.

A 50° lag would mean a far higher no-load power factor than a transformer core gives. The current cannot lead, since the transformer on no load is an inductive load.